【LeetCode】从尾到头反过来返回每个节点的值（用数组返回）day03

示例 1：
输入：head = [1,3,2]
输出：[2,3,1]
限制：
0 <= 链表长度 <= 10000

//使用栈的特性先进后出
//复杂度O（n）
public int[] reversePrint(ListNode head) {
        Stack<ListNode> stack =  new Stack<ListNode>();
        ListNode temp = head;
        while(temp!= null){
           stack.push(temp);
           temp = temp.next;
           }
           int[]  result =  new int[stack.size()];
           for(int i = 0 ;  i< stack.size();i ++ ){
           result[i] = stack.pop().val;
           }
        return result;
          
    }

//自己脑瓜子有点问题
    public int[] reversePrint(ListNode head) {
        ListNode[] result = new ListNode[];
        ListNode temp = head;
        int count = 0;
        while(temp != null ){
            result[count] = temp;
            temp = temp.next;
            count++;
        }
        int[] resultV2 = new int[count];
        for(int i = 0; i<count; i++){
            resultV2[count] = result[i].val;
            count--;
            }
            return resultV2;
}

//纯O（n）的复杂度
class Solution {
    public int[] reversePrint(ListNode head) {
        ListNode temp = head;
        int count = 0;
        while(temp != null ){
            temp = temp.next;
            count++;
        }
        int[] resultV2 = new int[count];
        temp = head;
        for(int i = count-1; i>=0; i--){
            resultV2[i] = temp.val;
            temp = temp.next;
        }
            return resultV2;
    }
}